Radius of convergence of a power series(complete result, hopefully)

[Susanne217]

Homework Statement

I have post this before but now I have come up with the complete result hopefully

Anyway given the power series $$\sum_{j=0}^{\infty} F_{j} z^j$$

find the radius of convergence around zero and $$F_j = F_{j-1} - F_{j-2}$$

and that $$j \geq 2$$

The Attempt at a Solution

I know that recursive relation $$F_j = F_{j-1} - F_{j-2}$$ can be rewritten to the polynomial $$z^ = z +1$$ which has the roots

$$z_{0},z_1 = \frac{1 \pm \sqrt{5}}{2}$$

and that I know from discrete mathematics that the general solution of $$F_j = A z_{0}^j + B z_{1}^j$$ and we know that if the series converges then

$$\lim_{n \to \infty} |\frac{F_{j+1}}{F_j}| < 1$$

thus by the ratio test

$$\lim_{n \to \infty} |\frac{F_{j+1}}{F_j}| = \lim_{n \to \infty} |\frac{\frac{\sqrt{5}+1}{2}^{n+1} A + \frac{\sqrt{5}+1}{2}^{n+1} B}{\frac{\sqrt{5}+1}{2}^{n} A+ \frac{\sqrt{5}+1}{2}^{n}B} | = z_0$$ and then

$$\lim_{n \to \infty} |\frac{F_{j+1}z^{j+1}}{F_j z^{j}}| = |z|z_0 = \frac{1+\sqrt{5}}{2}|z| < 1$$ and hende the radius of Convergence of the series around zero is

$$R = \frac{1+\sqrt{5}}{2}$$

How is that friends?

Susanne

 

[vela]

Homework Statement

I have post this before but now I have come up with the complete result hopefully

Anyway given the power series $$\sum_{j=0}^{\infty} F_{j} z^j$$

find the radius of convergence around zero and $$F_j = F_{j-1} - F_{j-2}$$

and that $$j \geq 2$$

The Attempt at a Solution

I know that recursive relation $$F_j = F_{j-1} - F_{j-2}$$ can be rewritten to the polynomial $$z^ = z +1$$ which has the roots

$$z_{0},z_1 = \frac{1 \pm \sqrt{5}}{2}$$

and that I know from discrete mathematics that the general solution of $$F_j = A z_{0}^j + B z_{1}^j$$ and we know that if the series converges then

$$\lim_{n \to \infty} |\frac{F_{j+1}}{F_j}| < 1$$

That's not correct. For example, if you had F_j=2^j and z=1/4, you'd get

$$\sum_{j=0}^\infty F_j z^j = 2^j 2^{-2j} = \sum_{j=0}^\infty 2^{-j} = 2$$

but |F_j+1/F_j| = 2 > 1. You can see that a series can converge even though the coefficients aren't bounded as long as z^j goes to zero fast enough.

Anyway, I think you're making this more complicated than it needs to be. If you plug in your solution for F_j into the original series, you get

$$\sum_{j=0}^\infty F_j z^j = \sum_{j=0}^\infty (A z_0^j +B z_1^j) z^j = A \sum_{j=0}^\infty (z_0 z)^j + B \sum_{j=0}^\infty (z_1 z)^j$$

Can you see where to go from there?

 

[Susanne217]

No I'm not sure please elaborate :)

I can see what you are saying that I made an error so that the series diverges. So I need to show that both parts of the sum convergence?

 

[vela]

Hint: The two series are geometric series.

 

[Susanne217]

Hint: The two series are geometric series.

Vela,

I have been looking in one of analysis books,

If I view the series as $$\sum_{n=0}^{\infty} F_n(z-z_0)^n$$ and let

$$\lim_{n \to \infty} sup |F_n|^{\frac{1}{n}}$$ and let $$r = \frac{1}{\lambda}$$

Where limit is $$+Inifity$$ and thusly by the theorem, r the radius of convergence is zero. Is this resonable argument here?

 

[Susanne217]

Hint: The two series are geometric series.

And I know Vela,

if I show that the two geometric series P1 and P2 converge on their own (Abel's theorem) then their sum

P1 + P2 converge as well...

I find $$A = \frac{1}{\sqrt{5}}$$ and $$B = -\frac{1}{\sqrt{5}}$$

and $$z_0, z_1 = \frac{1 \pm \sqrt{5}}{2}$$

But what I don't get Vela is that $$\lim{j \to \infty} \sum_{j = 0}^{\infty} (\frac{1 + \sqrt{5}}{2}z)^j$$

then for P1 being a geomtric series to converge if 0 < |z| < 1

thereby $$P1 = \frac{1}{\sqrt{5}} \sum_{j = 0}^{\infty} (\frac{1 + \sqrt{5}}{2}z)^j = \frac{\frac{1 + \sqrt{5}}{2}}{1-z} = \frac{1+\sqrt{5}}{2(z-1)}$$

But what I don't get Vela is that $$\lim_{j \to \infty} ( (\frac{1 + \sqrt{5}}{2})^j|z| =0)$$

 

[vela]

If I view the series as $$\sum_{n=0}^{\infty} F_n(z-z_0)^n$$ and let

$$\lim_{n \to \infty} sup |F_n|^{\frac{1}{n}}$$ and let $$r = \frac{1}{\lambda}$$

Where limit is $$+\infty$$ and thusly by the theorem, r the radius of convergence is zero. Is this a reasonable argument here?

You evaluated the limit incorrectly. Let $z_\pm = (1\pm\sqrt{5})/2$. You have

$$|F_n|^{\frac{1}{n}} = |Az_+^n+Bz_-^n|^{1/n} = z_+|A|^{1/n}\left|1+\frac{B}{A}\left(\frac{z_-}{z_+}\right)^n\right|^{1/n}$$

assuming A is not zero. You should be able to show that the lim sup is z₊.

 

[Susanne217]

You evaluated the limit incorrectly. Let $z_\pm = (1\pm\sqrt{5})/2$. You have

$$|F_n|^{\frac{1}{n}} = |Az_+^n+Bz_-^n|^{1/n} = z_+|A|^{1/n}\left|1+\frac{B}{A}\left(\frac{z_-}{z_+}\right)^n\right|^{1/n}$$

assuming A is not zero. You should be able to show that the lim sup is z₊.

okay don't I need to take the limit still?

 

[vela]

Yes, and the sup as well. I was just rewriting |F_n|^(1/n) in a way to make it easier to analyze when determining the supremum and the subsequent limit of the sequence of sups.

 

[vela]

I find $$A = \frac{1}{\sqrt{5}}$$ and $$B = -\frac{1}{\sqrt{5}}$$

I'm not sure what you did here. A and B are arbitrary constants. Unless you have additional info, like what F₀ and F₁ are, you can't determine them.

But what I don't get Vela is that $$\lim_{j \to \infty} \sum_{j = 0}^{\infty} (\frac{1 + \sqrt{5}}{2}z)^j$$

then for P1 being a geomtric series to converge if 0 < |z| < 1

thereby $$P1 = \frac{1}{\sqrt{5}} \sum_{j = 0}^{\infty} (\frac{1 + \sqrt{5}}{2}z)^j = \frac{\frac{1 + \sqrt{5}}{2}}{1-z} = \frac{1+\sqrt{5}}{2(z-1)}$$

But what I don't get Vela is that $$\lim_{j \to \infty} ( (\frac{1 + \sqrt{5}}{2})^j|z| =0)$$

To be honest, I have no idea what you're doing here. One thing I'll point out, though, is that the ratio of the geometric series isn't simply z.

 

[Susanne217]

You evaluated the limit incorrectly. Let $z_\pm = (1\pm\sqrt{5})/2$. You have

$$|F_n|^{\frac{1}{n}} = |Az_+^n+Bz_-^n|^{1/n} = z_+|A|^{1/n}\left|1+\frac{B}{A}\left(\frac{z_-}{z_+}\right)^n\right|^{1/n}$$

assuming A is not zero. You should be able to show that the lim sup is z₊.

Maybe I am a bit behind but $$z_+$$ is larger than 1?

 

[Susanne217]

I'm not sure what you did here. A and B are arbitrary constants. Unless you have additional info, like what F₀ and F₁ are, you can't determine them.

To be honest, I have no idea what you're doing here. One thing I'll point out, though, is that the ratio of the geometric series isn't simply z.

I'm told that $$F_0 = F_1 = 1$$

 

[vela]

Yes, z₊ is approximately 1.618 and z_- is approximately -0.618. What's more important is that |z_-/z₊|<1.

 

[Susanne217]

Yes, z₊ is approximately 1.618 and z_- is approximately -0.618. What's more important is that |z_-/z₊|<1.

But radius of convergence which I'm suppose to end up with is still what some books call the golden ratio?

 

[vela]

z₊ is the golden ratio. The radius of convergence will be its reciprocal.

 

[Susanne217]

z₊ is the golden ratio. The radius of convergence will be its reciprocal.

Thanks now I know what I am suppose to end up with :)

I will with the major hints you have given me re-do the calculations and then post them again then I get home for your (hopefully) approval :)

 

[Gib Z]

I approached this differently and seemed to have got a result different to both of you. In particular, I find that the Radius of Convergence is 1, which is greater than the reciprocal of the Golden Ratio (~0.618).

We start with the definition for the coefficients: $$F_0 = F_1 = 1, F_n = F_{n-1} - F_{n-2}$$

From this we have for n=0,1,2, F = 1, 1, 0 respectively.

Also: $$F_n = F_{n-1} - F_{n-2} = ( F_{n-2} - F_{n-3} ) - F_{n-2} = - F_{n-3}$$. So we can see the sequence for F is : 1 , 1, 0, -1, -1, 0 , 1, 1, 0, -1, -1, 0...

Hence our series certainly converges at least where $\sum_{p=0}^{\infty} x^p$ converges, that is, all x such that |x|< 1. But our series certainly doesn't converge if |x|= 1 or |x|>1, as the n-th term does not converge to zero.

 

[vela]

D'oh! It appears both Susanne and I made the same mistake solving the recurrence relation for F_j. Using the correct solution, I find a radius of convergence of 1 as well.

 

[Susanne217]

D'oh! It appears both Susanne and I made the same mistake solving the recurrence relation for F_j. Using the correct solution, I find a radius of convergence of 1 as well.

I get that solution to that version of the resursive relation to be -1? Is that correct then? This ruins everything now I get produce a general solution for F_j and I am totally lost on how you guys suddenly can conclude that the radius is insteed 1?

By the way what's not clear to me how can suddenly out of the blue conclude you must rewrite the recursive relation to $$F_j= - F_{j-3}$$ where did that come from?

 

[Gib Z]

I derived it in the original post. I wrote out the relation: F_n = F_(n-1) + F_(n-2)

Then I replaced F_(n-1) by the equivalent expression given by the same rule: F_(n-1) = F_(n-2) - F_(n-3)

Then I simplified. With that result, since we know the first 3 terms of the sequence, the rest come about easily and we can see why F is given by the sequence I gave in my previous post.

 

[Susanne217]

I derived it in the original post. I wrote out the relation: F_n = F_(n-1) + F_(n-2)

Then I replaced F_(n-1) by the equivalent expression given by the same rule: F_(n-1) = F_(n-2) - F_(n-3)

Then I simplified. With that result, since we know the first 3 terms of the sequence, the rest come about easily and we can see why F is given by the sequence I gave in my previous post.

Are you a 100% sure that is applicable here? Cause we are dealing with as I wrote j or n as you wrote which are larger than 2? and the original relation satisfies positive fibunacci numbers why re-write and deal with the negative once as well?

I would very appriacate if you would elaborate on why this re-write is needed?

and why is your approach then more correct than mine?

 

[Gib Z]

The original relation does NOT satisfy the Fibonacci numbers, which require F_0 = F_1 = 1, and F_n = F_(n-1) + F_(n-2). For YOUR question, there is a minus sign, not plus.

For our sequence F_n, which is not the Fibonacci sequence, we can show that F_n = -F_(n-3) if n is equal to or bigger than 3. So by the first post, we can derive the sequence F_n.

That sequence allows us to know what our power series actually is, and eventually allows us to find the Radius of Convergence for it.

 

[Susanne217]

The original relation does NOT satisfy the Fibonacci numbers, which require F_0 = F_1 = 1, and F_n = F_(n-1) + F_(n-2). For YOUR question, there is a minus sign, not plus.

For our sequence F_n, which is not the Fibonacci sequence, we can show that F_n = -F_(n-3) if n is equal to or bigger than 3. So by the first post, we can derive the sequence F_n.

That sequence allows us to know what our power series actually is, and eventually allows us to find the Radius of Convergence for it.

Okay,

Sorry I was a bit tired. If I change the original relation to F_n = F_(n-1) + F_(n-2) insteed of F_n = F_(n-1) + F_(n-2) will my original calculations on the radius of covergence fit? That the radius of convergence is the reciprocal of the golden ratio?

 

[Gib Z]

Did you mean if you change the original relation to F_n = F_(n-1) + F_(n-2) instead of F_n = F_(n-1) - F_(n-2) ?

If the coefficients of the power series were the Fibonacci numbers, then the radius of convergence would be the reciprocal of the golden ratio, yes.

 

[Susanne217]

Did you mean if you change the original relation to F_n = F_(n-1) + F_(n-2) instead of F_n = F_(n-1) - F_(n-2) ?

If the coefficients of the power series were the Fibonacci numbers, then the radius of convergence would be the reciprocal of the golden ratio, yes.

yes I mean that and it still satisfies that F(0) = F(1) = 1 and n>= 2...

 

[vela]

I get that solution to that version of the resursive relation to be -1? Is that correct then? This ruins everything now I get produce a general solution for F_j and I am totally lost on how you guys suddenly can conclude that the radius is insteed 1?

The characteristic polynomial of the original recursion relation, z²-z+1, doesn't have -1 as a root, but it does have complex roots which turn out to have a modulus of 1, which is why the radius of convergence turns out to be 1.

 

[Susanne217]

The characteristic polynomial of the original recursion relation, z²-z+1, doesn't have -1 as a root, but it does have complex roots which turn out to have a modulus of 1, which is why the radius of convergence turns out to be 1.

thanks, get it know :)